What is dy/dx for y - ln(1/y) + ln(y) = ln(x).
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calendarEducator since 2008
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y- ln (1/y) + ln y = ln x.
First we will rewrite ln (1/y) = ln y^-1 = - ln y.
==> y + ln y + ln y = ln x.
==> y + 2*ln y = ln x.
Now we will differentiate with respect to x.
==> (y)' + (2ln y)' = (ln x) '
==> y' + 2*1/y * y' = 1/x
==> y' + (2/y) y' = 1/x
Now we will factor y':
==> y' ( 1+ 2/y) = 1/x
==> y' ( y+2)/y = 1/x
Now we will multiply both sides by y/(y+2)
==> y' = 1/x * ( y/(y+2)
==> y' = y/x(y+2)
Then, dy/dx = y/x(y+2)
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We have y - ln ( 1/y) + ln y = ln x
use the relation that log a + log b = log a*b
=> y - ln [( 1/y)(1/y)] = ln x
=> y - ln [ 1/y^2] = ln x
=> dy/dx - y^2*(-2)*(1/y^3)*dy/dx = 1/x
=> dy/dx [ 1 + 2* y^2 / y^3] = 1/x
=> dy/dx [ 1 + 2/y]= 1/x
=> dy/dx = 1/[x*(1 + 2/y)]
The required value of dy/dx = 1/[x*(1 + 2/y)]
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